The equation for the displacement of a particle executing `SHM` is `x = 5sin (2pit)cm` Then the velocity at `3cm` from the mean position is( in cm//s)

Options

A. `2pi`
B. `3pi`
C. `4pi`
D. `8pi`

Explanation

To find the velocity of a particle executing Simple Harmonic Motion (SHM) when it is 3 cm from the mean position, we can follow these steps:

### Step 1: Identify the parameters from the displacement equation
The given displacement equation is:
\[ x = 5 \sin(2\pi t) \]
From this equation, we can identify:
- Amplitude \( A = 5 \) cm
- Angular frequency \( \omega = 2\pi \) rad/s

### Step 2: Use the formula for velocity in SHM
The formula for the velocity \( v \) of a particle in SHM is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where:
- \( A \) is the amplitude,
- \( x \) is the displacement from the mean position.

### Step 3: Substitute the known values
We want to find the velocity when the particle is at \( x = 3 \) cm. Substituting the values into the formula:
- \( A = 5 \) cm
- \( x = 3 \) cm
- \( \omega = 2\pi \) rad/s

The equation becomes:
\[ v = 2\pi \sqrt{5^2 - 3^2} \]

### Step 4: Calculate the expression inside the square root
First, calculate \( A^2 - x^2 \):
\[ A^2 = 5^2 = 25 \]
\[ x^2 = 3^2 = 9 \]
Thus:
\[ A^2 - x^2 = 25 - 9 = 16 \]

### Step 5: Calculate the velocity
Now substitute back into the velocity equation:
\[ v = 2\pi \sqrt{16} \]
\[ v = 2\pi \times 4 \]
\[ v = 8\pi \]

### Step 6: Approximate the value of velocity
Using \( \pi \approx 3.14 \):
\[ v \approx 8 \times 3.14 = 25.12 \text{ cm/s} \]

### Final Answer
The velocity of the particle when it is 3 cm from the mean position is approximately:
\[ v \approx 25.12 \text{ cm/s} \]

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